Question

the sum of a number and its square is 63the sum of the sum of a number and its square is 63 by 4 find the number ​

Answer 1

Let the number be 'x

According to the given condⁿ:

$x + {x}^{2} = \frac{63}{4} \\ {x}^{2} + x = \frac{63}{4} \\ solving \: by \: completing \: square \: method - \\ we \: need \: to \: add \\ {( \frac{1}{2} \times coefficient \: of \: x)}^{2} on \: both \: sides \\ = {( \frac{1}{2} \times 1)}^{2} = \frac{1}{4} \\ \\ adding \: \frac{1}{4} on \: both \: sides \\ \\ {x}^{2} + x + \frac{1}{4} = \frac{63}{4} + \frac{1}{4} \\ ( {x})^{2} + 2(x)( \frac{1}{2} ) + {( \frac{1}{2} )}^{2} = \frac{64}{4} = 16 \\ {(x + \frac{1}{2} )}^{2} = 16 \\ x + \frac{1}{2} = |16| = + or - 16 \\ \\ x + \frac{1}{2} = + 16 = > x = 16 - \frac{1}{2} = \frac{31}{2} \\ or \\ x + \frac{1}{2} = - 16 = > x = - 16 - \frac{1}{2} = \frac{ - 33}{2}$

Hence, the number is either 31/2 or -33/2.